Let $n$ be a positive integer. A Japanese triangle consists of $1+2+\cdots +n$ circles arranged in an equilateral triangular shape such that for each $i=1,2,\dots ,n$, the $i$-th row contains exactly $i$ circles, exactly one of which is coloured red. A ninja path in a Japanese triangle is a sequence of $n$ circles obtained by starting in the top row, then repeatedly going from a circle to one of the two circles immediately below it and finishing in the bottom row.

In terms of $n$, find the greatest $k$ such that in each Japanese triangle there is a ninja path containing at least $k$ red circles.

Step 1 (Upper bound construction): Place red circles to minimize the guaranteed number on any ninja path. The optimal adversarial strategy places red circles so that at each level, the red circle 'splits' the remaining paths evenly, ensuring each path hits few reds.

Step 2 (Binary tree structure): A ninja path is a root-to-leaf path in a binary tree of depth $n$. The red circles correspond to choosing one node per level. The question is: what's the maximum $k$ such that every choice of one-per-level hits every root-to-leaf path at least $k$ times?

Step 3 (Lower bound via pigeonhole): At each level $i$, the red circle is in one of $i$ positions. A ninja path passes through one specific position per level. If the red is at position $j$ in row $i$, the paths passing through position $j$ in row $i$ collect this red. The number of paths through position $j$ in row $i$ is $\left(\genfrac{}{}{0ex}{}{n-i}{\lfloor (n-i)/2\rfloor }\right)$-like.

By a halving argument: the adversary can avoid at most half the remaining paths at each step, so after $n$ levels, the minimum number of reds on any path is $\lfloor {\log }_{2}n\rfloor +1$.

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